Interest Compound Daily Calculator - PDF

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3 Approximating compound interest

The Simple Interest Formula (1.1.6) is just that: simple. As long as you remember the Period warning (1.1.13)
and Interest rate warning (1.1.9) and you just multiply and it is pretty hard to go wrong. What’s more there is
an easy way to check any answer: since changing the periods used to measure the term has no eﬀect on the ﬁnal
value of the interest, you can just recalculate with diﬀerent periods. The Compound Interest Formula (1.2.4) is
not really more complicated. However, since the choice of the compounding period now has a substantial eﬀect
on the answer, you can’t just blindly recalculate with diﬀerent periods to check. In this section, we’ll learn how
to keep our eyes open while changing periods. This leads to several easy approximations for compound interest
which can be used to check that a compound interest answer at least “looks right”. While doing this, we’ll ﬁnd out
that calculators don’t always compute the compound interest formula correctly and learn how to work around
their limitations. In later sections ( Yields (1.4) and Inﬂation, population, computation, radiation (1.5), for
example), what we’ll learn will turn out to be handy in many other ways.

The simple interest approximation

The ﬁrst approximation we’ll use is truly simple: just ignore compounding and use the Simple Interest For-
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mula (1.1.6). If we use years as periods — so p =             and T = y — then the interest on an amount A0 is
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I = p · A0 · T =     · A0 · y and the future value AT — the total of principal plus interest at the end of the period
100
r                      r
— is roughly A0 +       · A0 · y = A0 (1 +     · y). We can use this to approximate the compound interest formula.
100                    100

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Here’s an exanple.

Example (1.3.1) Let’s try checking my answer to one of the Problems in Compound interest (1.2). In part
b) of Problem (1.2.10) which asked for the future value S of \$2,600 at nominal interest of 9% for a term of
3 years compounded quarterly, my answer was \$3, 395.73. The Simple Interest Approximation (1.3.3) gives
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A0 (1 +     · y) = \$2600(1 + .09 · 3) = \$3, 302.00 which is just a bit smaller than my answer.
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If we think of A0 as an initial amount or present value B and think of AT as a ﬁnal amount or future value S, we
can rewrite the approximation
r                                 r
AT     A0 (1 +       · y)      as    S     B(1 +       · |y|)
100                               100
This new formula on the right can be used to check both future and present value problems. Just remember that
S is to be the ﬁnal amount and B the starting amount, so in a present value problem we’d have to reverse the A’s:
S = A0 and B = AT . The bars around the y are absolute value signs and remind us to always make y positive in
this formula — just as we have always done when working with simple interest.

Example (1.3.2) In part a) of Problem (1.2.16) which asked for the present value of a sum of S = \$4200 which
will be due in 4 years at a nominal interest rate of 12% compounded annually, I got B = \$2669.18. The Simple
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Interest Approximation (1.3.3) gives B(1 +        · |y|) = \$2669.18(1 + .12 · 4) = \$3950.39. There are two points
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to note. First, this time we plugged in the answer as the present value or starting amount — because we were
checking a present value problem. Second, even though I used a negative period of −4 years in Problem (1.2.16)
I plugged in |y| = 4 in the check.

Unfortunately, this “approximation” is only close to the correct value when the term is quite short. If you look

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over the example at the start of section Compound interest (1.2), you’ll see that over terms of many years, the
interest on the interest in a compound interest calculation can become much larger than the original amount
or the interest on the original amount. However, we can still make some use of this formula by thinking of it
somewhat diﬀerently. Using the Simple Interest Formula (1.1.6) amounts to making the entire term a single
period so that, in eﬀect, there is no compounding. In other words we want T = 1. We can check this idea by
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working work backwards. Since T = m · y by the Term Conversion Formula (1.1.14), m · y = 1 so m = . Then,
y
by the Interest Rate Conversion Formula (1.1.11)
r        r                  r
p=            =                 =       ·y.
100 · m   100 ·
1
y
100

Finally, applying the Compound Interest Formula (1.2.4)
r                   r
AT = A0 (1 + p)T = A0 (1 +            · y)1 = A0 (1 +     · y) .
100                 100
By combining this with what we know about the Effect of More Frequent Compounding (1.2.19), we can
squeeze out a bit of information even when the approximation is way oﬀ. Remember that more frequent com-
pounding increases future values and decrease present values. Since using simple interest amount to doing the
least compounding possible — none at all — it should deﬁnitely underestimate future values. Note that both
examples above conﬁrm this: the approximations are both slightly smaller than the answers being checked.

Simple Interest Approximation (1.3.3) The ending or future amount S in a compound interest calculation
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should be greater than the approximation B(1 +          · |y|) (where B is the starting or present value) and the
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approximation should be fairly good if the term is not too long.

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Both Example (1.3.2) and Example (1.3.2) illustrate reasonable uses of the Simple Interest Approximation (1.3.3).
I often do something even cruder in my head when working problems in class. In the ﬁrst case, I’ll say: “The
interest is .09 · 3 = .27 which is about 1 and 1 of \$2600 is \$650 so I should expect my answer to be a bit bigger
4     4
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than \$3250”. Similarly, in the second problem, I’d say, “Here the interest is .12 · 4 = .48 which is about 2 so I
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expect \$2669.18 times (1 + 2 ) = 2 to be a bit smaller then \$4200; it’s easier to turn this around 3 of \$4200 or
\$2800 should be somewhat bigger than \$2669.18”. The point is that since we are only approximating anyway we
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can aﬀord to ignore the diﬀerence between .27 and 4 or between .48 and 2 . I know I am much more likely to
perform a quick mental check than one where I have to get out my calculator and a check you don’t perform is
not much of a check.

Example (1.3.4) To see the limitations of the approximation, let’s look at part c) of Problem (1.2.17) which
asked for the present value of an amount of \$100,000 due in 40 years at an interest rate of 4.8% compounded
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monthly. My answer was \$14, 716.95. The Simple Interest Approximation (1.3.3) gives B(1 +                · y) =
100
\$14, 716.95(1 + .048 · 40) = \$42.973.49. Again this is less than the correct futue value of \$100,000, but now it
is so much less — barely two-ﬁfths — that is not much use as a check. The period here was just too long for the
approximation to be useful.

Self-Test

Problem (1.3.5) Use the Simple Interest Approximation (1.3.3) to check your answers to Problem (1.2.11),
Problem (1.2.21) and Problem (1.2.23). Are there any other problems in Compound interest (1.2), for which
you’d expect the Simple Interest Approximation (1.3.3) to be fairly accurate?

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The continuous approximation

We have already seen in Effect of More Frequent Compounding (1.2.19) that keeping the nominal rate and
the term in years ﬁxed, the more often we compound the larger the amount owed at the end of the term. Let’s
go back to the Problem (1.2.2) where we borrowed \$100,000 at 8% interest and ask: What happens if the bank
compounds ever more frequently? Let’s try compounding monthly, daily, hourly and once a second in the two
problems above. The corresponding vales of m are : 12, 365, 8760 and 31,536,000. (In other words, there are
8760 hours and 31,536,000 seconds in a year. So while a million seconds might seem like forever it is actually
only about 11.57 days.) Now we can just apply the Method for ﬁnding compound interest (1.2.15) —- I have
omitted the calculations — to get Table (1.3.6) below for a term of 3 years:

m         12                  365                     8760       31,536,000
T         36                 1095                    26280       94,608,000
p        0.00˙
6        .2191780822e-3          .9132420088e-5   .2536783358e-8
AT     \$127,023.71       \$127,121.56              \$127,123.37    \$132,819.91

Table (1.3.6) Compounding table for a 3 year term

and Table (1.3.7) for a term of 12 years.

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m         12                   365                     8760       31,536,000
T         144                 4380                   105120      378,432,000
p            ˙
0.006         .2191780822e-3          .9132420088e-5   .2536783358e-8
AT     \$260,338.92        \$261,142.08              \$261,156.97    \$311,209.46

Table (1.3.7)     Compounding table for a 12 year term

Problem (1.3.8) Make your own calculation of each of the amounts in the tables above. You should get the
answers in the table to the penny when you compound monthly. Some of your other answers may be somewhat
diﬀerent for reasons I’ll explain. If so, don’t worry.

There are many interesting things to note about these tables. Let’s look at the p rows to start. First notice that
these rows in the two tables contain identical values: we should expect this since p does not depend on the term
of the loan — which is what diﬀers between the two tables — but only on the frequency with which we compound.
Second, notice how tiny the periodic rates have become: this is because we are compounding many times a year
with very short compounding periods and hence we get very little interest in each period. The values are so small
that my calculator has given them to me in scientiﬁc notation so I will have as many decimals as possible. Recall
that a value like .2536783358e − 8 stands for .2536783358 × 10−8 = 0.000000002536783358 and my calculator
screen just does not have room for all those 0’s. Notice also that I did not round this number to preserve the

Next, let’s look at the ﬁnal amount of AT rows. Does anything strike you about these? All the amounts are getting

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closer and closer together — in the three year table they seem to be settling down around \$127,120 or so and
in the 12 year table around \$261,150 — and then suddenly the last amount where we compound in seconds is
much bigger. What’s going on? Two things. First, the ﬁnal amounts are wrong! The problem is that to ﬁnd them
I asked my calculator to compute

AT = 100000 ∗ (1 + .2536783358e − 8)94,608,000            and    AT = 100000 ∗ (1 + .2536783358e − 8)378,432,000

and those exponents made it choke. It just can’t compute powers that big accurately. The 10 digits of precision
it uses is just not enough to get the ﬁnal amount even to 2 places!! If I use a much better calculator (which carries
20 places) and make the same calculation, I ﬁnd that the amounts turn out to be

AT , 3 years   \$127,023.71         \$127,121.56        \$127,124.78   \$127,124.92
AT , 12 years   \$260,338.92         \$261,142.18        \$261,168.50   \$261,169.65

Table (1.3.9) Frequent compounding table computed with a 20 place calculator

Notice that it turns out it wasn’t just the ﬁnal “seconds” amounts that were wrong. They were just the only wrong
answers that were so far oﬀ that it was clear to the naked eye that something was ﬁshy. Both the amounts for
hourly compounding were oﬀ by dollars and the 12 year ﬁnal amount with daily compounding — which is what
your bank uses — was oﬀ by 10 cents. But all these other answers were close enough that the only way we would
ever know that they were wrong was by making a second more accurate calculation. The moral here is:

Murphy’s Law of Calculators (1.3.10)               Never trust a calculator’s answers unless you have some other way
to check them.

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In particular, we have now learned: Never use a calculator to take a power with a very large exponent. You may
be wondering how you are going to do problems like the one where we got the wrong answer if you are not able
to use your calculator. Relax: I’m just not going to ask you to work any problem where the exponent in the
Compound Interest Formula (1.2.4) is dangerously large. But, in real life where daily compounding is common,
you might be asked to: if you are, remember to watch out.

Did some of the amounts you computed in Problem (1.3.8) diﬀered from mine as I suggested in the problem that
and (unless you have a very good calculator) that yours probably would be wrong too. But there’s no chance we’d
get the same wrong answers. Why? Because every calculator is a bit diﬀerent inside. While they’ll all give the
same answer when they can get the right one, when things go wrong each calculator goes wrong in its own way.

So much for the bad news. Let’s get back to those amounts. The correct answers give a striking conﬁrmation of
our initial impression that as you compound more and more frequently, the ﬁnal amounts get larger and larger
but do so ever more slowly. Eventually, these amounts appear to settle down. In fact, no matter how often you
compound — even if you compound a trillion times a second — the ﬁnal amount you’ll wind up with in these
two problems will never grow by another cent: after 3 years, you’ll have \$127,124.92 and after 12 you’ll have
\$261,169.65. (You need a calculator that keeps 25 places to check these answers so you’ll just have to trust me
on this. If this worries you a bit, take a gold star: you’re catching on.)

What would be very nice is to have some “easy way” to get this magic maximum amount. It’s not that we’d ever
want to compound interest every second in real life and so have a direct need for a way around the limitations of
our calculators. But, if we did we’d have a good way to make a rough check of any compound interest calculation.
Our answer should be close to, but somewhat smaller than this magic amount: the more frequently we compound,

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the closer the two should be. In particular when we compound monthly or daily as in the majority of real world
loans we should see a few matching digits, as in the tables above. If we do not, we’ll know right away that
something is wrong.

Well kids, life is good! Finding formulas for limits – which is fancy way to say, for how things “settle down” —
is one of the main applications of calculus. And, a standard formula from calculus computes the magic amount
which appears in the tables above. Moreover, you don’t need to know any calculus to understand and use this
formula.

Continuous Approximation (1.3.11) AT lies between A0 and the continuous approximation
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A0 · e(p·T ) = A0 · e    100 ·y   .

If the compounding period is short, AT is close to the continuous approximation.

The Number e (1.3.12) The base e in this formula is a very important number: e 2.71828182845904523536.
But you don’t ned to try to memorize any of these decimals: e is so important it’s got its very own key on your
calculator. Moreover, exponentials with base e occur so often in so many places that there is also a key usually
called exp for taking the exponential base e of the current value.
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So to use the Continuous Approximation (1.3.11), you just calculate the product      · y, hit exp and multiply
100
by the amount A and you’ve got the magic amount. In the two problems above, it gives
r                                                                 r
A·e   100 ·y   = \$100000 · e.08·3 = \$127, 124.92         and     A·e    100 ·y    = \$100000 · e.08·12 = \$261, 169.65 .

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You might wonder why I preferred to calculate the exponent in the form            · y rather than in the apparently
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simpler form p · T . First, let’s remark that they really are equal. Using the Interest Rate Conversion For-
r              r
mula (1.1.11) and Term Conversion Formula (1.1.14) gives p · T =                m·y =         · y. Forgetting to use
100 · m          100
one of these conversion formulas is the most common error in working interest problems. Thus, the fact that I
can use the nominal interest rate and the term in years in the formula with no need for converting to periodic
rates and periods, makes the continuous approximation perfect for catching such errors. It’s one of the nicest
features of the formula that it lets us work with the real life quantities we like to think in terms of — nominal
rates and years.

But the formula has other amazing properties. First, even with a standard calculator you can use it to ﬁnd the
magic number to the penny. Using the Compound Interest Formula (1.2.4), I needed a supercalculator to get
these numbers. You’d have no way to compute them. Makes you think there might be something to this calculus
after all, and that it might not be as hard as it’s cracked up to be. (Both guesses are correct and I hope this
section will inspire a few of you to take a calculus course. If you are planning to do serious work in any of the
mathematical, computational, physical, biological or social sciences, you will have to do so eventually and the
sooner you start the easier a time you’ll have with the math and the further ahead you’ll be in your major. If you

The ﬁnal remarkable feature of the Continuous Approximation (1.3.11) is that the signs of the quantities T
and y which measure time are not mentioned anywhere. So far we have only used the formula in future value
problems where these quantities are positive but everything works just as well in present value problems when
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they are negative. The future value version says that when T and y are positive that A0 < AT < A0 e           100 ·y   or,

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since AT = A0 (1 + p)T that
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A0 < A0 (1 + p)T < A0 e             100 ·y

. The present value version says that if we replace T by −T and y by −y we should have
r
A0 > A0 (1 + p)−T > A0 e           100 ·(−y)    .

Note that, although the directions of the inequalities are reversed, the exact value AT is still in the between A0
and the continuous approximation as claimed. I leave this to you: it is good practice in playing with exponents
and inequalities: you need no special knowledge about the number e.
r                                                  r
Problem (1.3.13) Show that if A0 < A0 (1 + p)T < A0 e                100 ·y   then A0 > A0 (1 + p)−T > A0 e             100 ·(−y)   .

If you are hoping that I will now explain where this approximation comes from, bless you. First oﬀ, there’s not
much doubt that the Continuous Approximation (1.3.11) is correct. The fact that it computes the two amounts
above to the penny is pretty convincing. And, as we’ve already noted you don’t need to understand where it
comes from to use it. So if you could care less, you can skip to Example (1.3.15).
N
1
Everything comes down to a set of approximations to the number e: if N is a large positive number, then 1 + N
is slightly smaller than e. The bigger N you take, the closer the approximation. You can convince yourself of this
by calculating with a few big values of N. For example,
1000                                                             10000
1                                                           1
1+                 = 2.716923932           and            1+                         = 2.718145927 .
1000                                                        10000

The ﬁrst less than e by about .013 and the second by about .0014.

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Problem (1.3.14) You shouldn’t take N too big, however. Why? What will happen if you do?

I’m afraid if you want to know where this formula comes from you’ll have to take that calculus course. But we
can easily see how it leads to the continuous approximation. If we take the (p · T )th power of both sides, we ﬁnd
that
(p·T )
1 N               1 (N·p·T )
1+                = 1+            is a bit smaller than e(p·T ) .
N                 N
The Continuous Approximation (1.3.11) just says we can always ﬁnd a value of N for which the exponential
1 (N·p·T )
1+              equals the exponential (1 + p)T in the Compound Interest Formula (1.2.4). Equating bases we
N
1
need     = p and equating exponents we need N · p · T = T : there are two equations for the single unknown N
N
which would usually be impossible to satisfy. However, here is where a small miracle happens. Solving the ﬁrst
1
equation tells that we must have N = . If we plug this into the left side of the second equation it becomes
p
1
N · p · T = · p · T = T so the exponents automatically match up too! The ﬁnal point to note is that to get a
p
good approximation we need to have N large. But
1        1         100 · m
N=      =      r     =           .
p      100·m
r

Thus N is big when m is: in other words, we get a good approximation when we are compounding frequently.

Any way, using this approximation is a cinch. Let’s use it to check a few problems from the last section.

Example (1.3.15)      Problem (1.2.11) asked about an amount of \$1,255,000 earning nominal interest of 6.73%

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for a term of 5 years. In part c), where we compounded monthly I got a future value of \$1,755,397.74. The
6.73
Continuous Approximation (1.3.11) gives \$1, 255, 000e             100 ·5   = \$1, 757, 048.78 which is, as predicted, slightly
higher but matches the exact answer to 3 places.

Example (1.3.16) In part b) of Problem (1.2.18), we computed the present value of \$100,000 due in 20 years
at interest of 9.6% compounded quarterly to be \$14,996.97. The Continuous Approximation (1.3.11) gives
9.6
\$100, 000e 100 ·(−20) = \$14, 660.70. Note that since this was a present value problem where we were moving
money backwards in time, we used a negative value y = −20 and that this time, as expected, the continuous
approximation is slightly lower than the exact answer. Because we are compounding less frequently here, we get
a less accurate approximation — only 2 places match. But, the approximation is good enough that we’d be sure
to catch any silly errors like forgetting to convert from rate or term or miskeying one of the numbers into our
calculator.

Self-Test

Problem (1.3.17) Use the continuous approximation to check your answers to the 4 problems Problem (1.2.20)
to Problem (1.2.23).

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